Sir I'm having a doubt in option B . As every regular language is also a CFL so intersection of L1 and L2 will imply intersection of two context free languages. as the CFLs are not closed under intersection, so this should not necessarily be a CFl. ??

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–4 votes

L1 : CFL , L2 : RL
option A : L1-L2
=L1∩(∼L2)
=CFL∩(∼RL)
=CFL∩(RL) ( as RL is closed under complementation )
=CFL∩CFL ( as if lang is RL then it is CFL also )
may be CFL or not ( as CFL is not closed under intersection)
hence option A statement may be false may not be ( here nothing mentioned about "ALWAYS")
option B : same logic as option A
hence option B statement may be false may not be
option C : CFL is not closed under intersection so we cant say anthing about it whether it is CFL or not
option D : True as RL is closed under complementation

but question asking about FALSE statements ( all given options are TRUE)so Ans is None